Answer: 441

Small Hint:

Set $t = \log_{2026} x$ and take $\log_{2026}$ of both sides to get a quadratic in $t$

Big Hint:

The two roots satisfy $t_1 + t_2 = 20,$ so the product of the solutions $x$ is $2026$ raised to that power; factor $2026$

Solution:

Let $t = \log_{2026} x.$ Taking $\log_{2026}$ of both sides of $x^{\frac{\log_{2026} x}{20}} = 26x$ gives $$\frac{t^2}{20} = \log_{2026} 26 + t,$$ that is $$t^2 - 20t - 20\log_{2026} 26 = 0.$$ The discriminant $400 + 80\log_{2026} 26$ is positive, so there are two real roots $t_1, t_2,$ each giving a valid positive solution $x = 2026^{t}.$

By Vieta’s formulas $t_1 + t_2 = 20,$ so the product of the solutions is $2026^{t_1} \cdot 2026^{t_2} = 2026^{20}.$ Since $2026 = 2 \cdot 1013$ and $1013$ is prime, $P = 2^{20} \cdot 1013^{20}$ has $21 \cdot 21 = 441$ positive divisors.